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JEE Main 8 Jan 2020 Shift 2 Solved Paper
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© examsnet.com
Question : 68
Total: 75
Let S be the set of all real roots of the equation,
3
x
(
3
x
−
1
)
+
2
=
|
3
x
−
1
|
+
|
3
x
−
2
|
.
Then S :
is an empty set.
contains at least four elements.
contains exactly two elements.
is a singleton.
Validate
Solution:
Let
3
x
=
t
;
t
>
0
t
(
t
−
1
)
+
2
=
|
t
−
1
|
+
|
t
−
2
|
t
2
−
t
+
2
=
|
t
−
1
|
+
|
t
−
2
|
Case-I :
t
<
1
t
2
−
t
+
2
=
1
−
t
+
2
−
t
t
2
+
2
=
3
−
t
t
2
+
t
−
1
=
0
t
=
−
1
±
√
5
2
t
=
√
5
−
1
2
is only acceptable
Case-II :
1
<
t
<
2
t
2
−
t
+
2
=
t
−
1
+
2
−
t
t
2
−
t
+
1
=
0
D
<
0
no real solution
Case-Ill :
t
>
2
t
2
−
t
+
2
=
t
−
1
+
t
−
2
t
2
−
3
t
5
=
0
⇒
D
<
0
no real solution
© examsnet.com
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